对多个python字典的对应元素求和 [英] Sum corresponding elements of multiple python dictionaries

问题描述

我有任意数量的等长python字典，带有匹配的键集，如下所示:

I have an arbitrary number of equal-length python dictionaries with matching sets of keys, like this:

{'a':1, 'b':4, 'c':8, 'd':9}

{'a':2, 'b':3, 'c':2, 'd':7}

{'a':0, 'b':1, 'c':3, 'd':4}
...

如何获得具有相同键集但值等于字典集中相应元素之和的单个字典?换句话说，我想要:

How can I obtain a single dictionary with the same set of keys but with values as the sums of corresponding elements in the dictionary set? In other words, I'd want:

{'a':3, 'b':8, 'c':13, 'd':20}

也许有一个丑陋，复杂的循环结构，但是有某种更好的方法来实现某种列表/字典理解技巧吗?想到它，我真的不确定如何制作一个丑陋的循环版本..

Maybe there's an ugly, complicated loop structure, but is there a nicer way to do this with some kind of list/dict comprehension cleverness? Come to think of it, I'm really not sure how to make an ugly loop version, anyway..

推荐答案

collections.Counter()进行救援；-)

from collections import Counter
dicts = [{'a':1, 'b':4, 'c':8, 'd':9},
         {'a':2, 'b':3, 'c':2, 'd':7},
         {'a':0, 'b':1, 'c':3, 'd':4}]
c = Counter()
for d in dicts:
    c.update(d)

然后:

>>> print c
Counter({'d': 20, 'c': 13, 'b': 8, 'a': 3})

或者您可以将其更改回字典:

Or you can change it back to a dict:

>>> print dict(c)
{'a': 3, 'c': 13, 'b': 8, 'd': 20}

Counter()是否所有输入字典都具有相同的键并不重要.如果您确定知道这样做的话，则可以尝试可笑的;-)这样的单线:

It doesn't matter to Counter() whether all the input dicts have same keys. If you know for sure that they do, you could try ridiculous ;-) one-liners like this:

d = {k: v for k in dicts[0] for v in [sum(d[k] for d in dicts)]}

Counter()更清晰，更快，更灵活.公平地说，这种稍微不太荒谬的单行代码也不太荒谬:

Counter() is clearer, faster, and more flexible. To be fair, though, this slightly less ridiculous one-liner is less ridiculous:

d = {k: sum(d[k] for d in dicts) for k in dicts[0]}

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